If $x \odot y = 4y+2$ and $x \oplus y = x^{2}+4y^{2}$, find $(4 \oplus 6) \odot 1$.
Solution: We don't need to find $4 \oplus 6$ because $x \odot y$ depends only on the right operand. Find $x \odot 1$ $ x \odot 1 = (4)(1)+2$ $ \hphantom{x \odot 1} = 6$.